Question

Find the direction ratios of a line perpendicular to both the lines whose direction ratios are 3, 2,

–1 and 2, 4, –2.​

Answer 1

Step-by-step explanation:

Given direction ratios are :−2,1,−1 and −3,−4,1

Let a,b and c be the direction ratios of the line perpendicular to the given lines.

Thus, we have,

−2a+b−c=0

−3a−4b+c=0

Cross multiplying, we get

1×1−(−4)×(−1)

a

=

(−3)×(−1)−(−2)×1

b

=

−2×−4−(−3)×1

c

⇒

1−4

a

=

3+2

b

=

8+3

c

⇒

−3

a

=

5

b

=

11

c

Let us find

a

2

+b

2

+c

2

=

(−3)

2

+5

2

+11

2

=

9+25+121

=

155

Thus, the direction ratios of the required line are −3,5,11

The direction cosines are :

155

−3

,

155

5

,

155

11

Answer 2

The direction ratios of a line perpendicular to both the lines whose direction ratios are 3, 2, –1 and 2, 4, –2 is 0,4,8.

Step 1 : Find the direction ratio.

The direction ratios of two lines 3,2,-1 and 2,4,-2 .

Let, $b_{1} = 3i +2j-k$

and $b_{2}=2i+4j-2k$

Let $b$ be the line perpendicular to given lines

b = $b_{1}$ ×$b_{2}$

$\left|\begin{array}{ccc}i&j&k\\3&2&-1\\2&4&-2\end{array}\right|$

= $i(-4+4)-j(-6+2)+k(12-4)$

= $0i +4j+8k$

= $4j+8k$

So, direction ratios = (0,4,8)